A sample of 300ml of a solution, magnesium chloraye Mg(ClO3)2, has a molar concentration if 4.02 mol/L. How much mass (grams) of magnesium coloration is in this sample?

1 answer

To find the mass of magnesium chlorate (Mg(ClO₃)₂) in the solution, we can use the formula:

\[ \text{mass} = \text{molarity} \times \text{volume} \times \text{molar mass} \]

  1. Molarity (C): Given as 4.02 mol/L.
  2. Volume (V): Given as 300 mL. We convert this to liters:

\[ 300 \text{ mL} = 0.300 \text{ L} \]

  1. Molar Mass (MM) of Mg(ClO₃)₂:
    • Magnesium (Mg): 24.305 g/mol
    • Chlorine (Cl): 35.453 g/mol
    • Oxygen (O): 16.00 g/mol

Calculating molar mass:

  • For Mg(ClO₃)₂: 1 Mg + 2 Cl + 6 O \[ \text{MM} = 1(24.305) + 2(35.453) + 6(16.00) \] \[ = 24.305 + 70.906 + 96.00 \] \[ = 191.211 \text{ g/mol} \]
  1. Calculate the mass: \[ \text{mass} = (4.02 \text{ mol/L}) \times (0.300 \text{ L}) \times (191.211 \text{ g/mol}) \] \[ = 4.02 \times 0.300 \times 191.211 \] \[ = 4.02 \times 57.3633 \] \[ = 230.47 \text{ g} \]

Thus, the mass of magnesium chlorate in the sample is approximately 230.47 grams.